Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 14

Expanding the determinant along the first row givesIn particular the matrix is invertible.

The cofactor matrix is $\left(\begin{array}{ccc}-3& -1& -1\\ 5& -1& 3\\ -1& 1& 1\end{array}\right)$. Transpose it to form the adjugate matrix, and divide by the determinant $-4$:Direct matrix multiplication gives $A{A}^{-1}=I$, checking all entries.

A homogeneous square linear system has a nonzero solution exactly when its coefficient matrix is singular. HereThusFor this value, the second equation gives $x=-2y$ and the first gives $z=y$; the third then holds automatically. All solutions are $t(-2,1,1)$, with $t\ne 0$ for the nonzero ones.

The coefficient matrix is a Vandermonde matrix, with determinantTherefore a nonzero solution exists exactly whenFor example if $a=b$, the first two columns are equal and $(1,-1,0)$ is a solution. If exactly two parameters agree, the distinct remaining column is independent of the repeated one, so the kernel has dimension one. If all three agree, only $x+y+z=0$ remains, giving a two-dimensional kernel.